Extremum problems in geometry and solids

Calculus, grade 11 (5 units) Lesson 19 of 26Learning

Extremum problems in geometry and solids

A problem in geometry

A problem in geometry usually has two changing quantities and one relation between them – a given radius, volume or area. So the solution starts with a drawing and the relation:

Draw the figure, and choose one variable, $\displaystyle x$ .
Express the second quantity by $\displaystyle x$ , from the given relation: Pythagoras’ theorem, the volume formula, the area formula.
Write the quantity to make greatest or least as a function of $\displaystyle x$ alone.
Decide the interval from the figure: a positive length, a point that stays on the circle or on the sheet.
Differentiate, solve $\displaystyle {f'(x)}={0}$ , and decide the kind of the point – by a table of signs, or by comparing with the ends of a closed interval.

Example: a rectangle in a semicircle

A rectangle is inscribed in a semicircle of radius 5: one side of it lies on the diameter, and the other two vertices are on the arc.
What is the greatest area the rectangle can have?

The solution below writes $\displaystyle x$ for the distance from the centre $\displaystyle O$ to the side edge of the rectangle, and expresses the height by Pythagoras’ theorem.

The vertex on the arc is 5 from $\displaystyle O$ , so the height of the rectangle is $\displaystyle \sqrt{25-x^2}$ , and the base is $\displaystyle 2x$ .
$\displaystyle {S(x)}={2x\sqrt{25-x^2}}$ , on the interval $\displaystyle 0<x<5$ .
By the product rule: $\displaystyle {S'(x)}={2\sqrt{25-x^2}+2x\cdot\frac{-2x}{2\sqrt{25-x^2}}}={\frac{2(25-x^2)-2x^2}{\sqrt{25-x^2}}}={\frac{50-4x^2}{\sqrt{25-x^2}}}$ .
$\displaystyle {S'(x)}={0}$ where $\displaystyle {x^2}={12.5}$ , and in the interval $\displaystyle {x}={\sqrt{12.5}}={\frac{5}{\sqrt{2}}\approx3.54}$ .
Before the point the numerator is positive, and after it negative: a maximum.
The height is $\displaystyle {\sqrt{25-12.5}}={\frac{5}{\sqrt{2}}}$ , and the greatest area is $\displaystyle {S}={2\cdot\frac{5}{\sqrt{2}}\cdot\frac{5}{\sqrt{2}}}={25}$ .

In the greatest rectangle the base is twice the height: it is made of two squares.

Example: a can with the least surface

A closed can in the shape of a cylinder must hold $\displaystyle 54\pi$ cm³.
What should its radius and height be so that its metal – its surface area – is as small as possible?

The solution below expresses the height by the radius $\displaystyle r$ from the volume, which gives a rational function of $\displaystyle r$ :

The volume: $\displaystyle {\pi r^2h}={54\pi}$ , so $\displaystyle {h}={\frac{54}{r^2}}$ .
The surface area: two bases and the curved surface, $\displaystyle {S}={2\pi r^2+2\pi rh}$ . Substituting $\displaystyle h$ : $\displaystyle {S(r)}={2\pi r^2+\frac{108\pi}{r}}$ , on the interval $\displaystyle r>0$ .
$\displaystyle {S'(r)}={4\pi r-\frac{108\pi}{r^2}}={0}$ , so $\displaystyle {r^3}={27}$ and $\displaystyle {r}={3}$ .

$\displaystyle r$	$\displaystyle 0<r<3$	$\displaystyle 3$	$\displaystyle r>3$
$\displaystyle S'(r)$	−	0	+
$\displaystyle S(r)$	falls	minimum	rises

The radius is 3 cm, and the height $\displaystyle {h}={\frac{54}{9}}={6}$ cm: the height equals the diameter.
The least surface area: $\displaystyle {S(3)}={18\pi+36\pi}={54\pi}$ cm².

Exercise 1

From a square sheet of card with side 30 cm, a square of side $\displaystyle x$ cm is cut out at each corner.
The edges are folded up, which makes an open box.

Show that the volume of the box is $\displaystyle {V(x)}={x(30-2x)^2}$ .
What is the interval of $\displaystyle x$ ?

Find $\displaystyle V'(x)$ .

$\displaystyle V(x)=x(30-2x)^{2}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {V'(x)}={0}$ .

Work out the greatest volume.

$\displaystyle V(x)=x(30-2x)^{2}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Which solution of the equation did you reject, and why?
Show that the point you chose is a maximum.

Exercise 2

A closed can in the shape of a cylinder holds $\displaystyle 16\pi$ cm³.
$\displaystyle r$ – the radius of the base, in cm.

Show that the surface area of the can is $\displaystyle {S(r)}={2\pi r^2+\frac{32\pi}{r}}$ .

Find $\displaystyle S'(r)$ .

$\displaystyle S(r)=2\pi r^{2}+\frac{32\pi}{r}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {S'(r)}={0}$ .

Work out the least surface area.

$\displaystyle S(r)=2\pi r^{2}+\frac{32\pi}{r}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

What is the height of this can?
Compare with the example: how are the height and the radius related?

The height

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The hypotenuse of a right-angled triangle is 6 cm.
$\displaystyle x$ – the length of one of the legs, in cm.
The area of the triangle is $\displaystyle {S(x)}={\frac{1}{2}x\sqrt{36-x^2}}$ .

Find $\displaystyle S'(x)$ .

$\displaystyle S(x)=\frac{1}{2}x\sqrt{36-x^{2}}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {S'(x)}={0}$ .

Work out the greatest area.

$\displaystyle S(x)=\frac{1}{2}x\sqrt{36-x^{2}}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

How long is the other leg in this triangle?
What triangle did you get?

Exercise 4

A lorry drives 500 km at a constant speed of $\displaystyle v$ km/h.
Every hour of driving costs 400 ₪ (the driver and the lorry), plus fuel at $\displaystyle \frac{v^2}{16}$ ₪ an hour.
The permitted speed: $\displaystyle 40\leq v\leq100$ .
The cost of the whole journey is $\displaystyle {C(v)}={\frac{500}{v}\left(400+\frac{v^2}{16}\right)}={\frac{200000}{v}+31.25v}$ ₪.

Find $\displaystyle C'(v)$ .

$\displaystyle C(v)=\frac{200000}{v}+31.25v$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {C'(v)}={0}$ .

Work out the cost at the point you found and at both ends of the interval.

At the point you found

$\displaystyle C(v)=\frac{200000}{v}+31.25v$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle C(40)

$\displaystyle C(v)=\frac{200000}{v}+31.25v$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle C(100)

$\displaystyle C(v)=\frac{200000}{v}+31.25v$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

At what speed is the journey cheapest, and what does it cost?
If the permitted speed were at most 70 km/h – at what speed would the journey be cheapest? Explain.

The cheapest speed

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The cost

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

At most 70 km/h

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: calculus, grade 11 (5 units)

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Rules of differentiation and the tangent line

The derivative of a power
$\displaystyle (x^n)'$ $\displaystyle =$ $\displaystyle nx^{n-1}$
The derivative of a product
$\displaystyle [f(x)\cdot g(x)]'$ $\displaystyle =$ $\displaystyle f'(x)\cdot g(x)+f(x)\cdot g'(x)$
The derivative of a quotient
$\displaystyle \left[\frac{f(x)}{g(x)}\right]'$ $\displaystyle =$ $\displaystyle \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$
The derivative of a composite function – the outer derivative times the inner derivative
$\displaystyle [f(u(x))]'$ $\displaystyle =$ $\displaystyle f'(u)\cdot u'(x)$
The derivative of $\displaystyle \frac{1}{x}$
$\displaystyle \left(\frac{1}{x}\right)'$ $\displaystyle =$ $\displaystyle -\frac{1}{x^2}$
The tangent line at the point $\displaystyle (x_1,y_1)$ on the graph
$\displaystyle y-y_1=f'(x_1)(x-x_1)$

The second derivative

The second derivative – the derivative of the derivative
$\displaystyle f''(x)$ $\displaystyle =$ $\displaystyle \left(f'(x)\right)'$
The function is concave up ( $\displaystyle \cup$ )
$\displaystyle f''(x)>0$
The function is concave down ( $\displaystyle \cap$ )
$\displaystyle f''(x)<0$
A possible inflection point – check that the sign of $\displaystyle f''$ changes there
$\displaystyle f''(x_0)=0$
A minimum point
$\displaystyle f'(x_0)=0,\ f''(x_0)>0$
A maximum point
$\displaystyle f'(x_0)=0,\ f''(x_0)<0$
Acceleration – the derivative of the speed
$\displaystyle a(t)=v'(t)=s''(t)$

Asymptotes

A vertical asymptote – $\displaystyle f(x)$ grows or falls without end as $\displaystyle x$ approaches $\displaystyle a$
$\displaystyle x=a$
A horizontal asymptote – $\displaystyle f(x)$ approaches $\displaystyle b$ as $\displaystyle x\to\infty$ or as $\displaystyle x\to-\infty$
$\displaystyle y=b$

Roots

The root of a square
$\displaystyle \sqrt{x^2}$ $\displaystyle =$ $\displaystyle |x|$
The derivative of $\displaystyle \sqrt{x}$
$\displaystyle (\sqrt{x})'$ $\displaystyle =$ $\displaystyle \frac{1}{2\sqrt{x}}$ $\displaystyle x>0$
The derivative of the root of a function
$\displaystyle (\sqrt{f(x)})'$ $\displaystyle =$ $\displaystyle \frac{f'(x)}{2\sqrt{f(x)}}$ $\displaystyle f(x)>0$

Radians and periods

Half a turn in radians
$\displaystyle \pi$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The length of an arc of $\displaystyle \alpha$ radians in a circle of radius $\displaystyle R$
$\displaystyle l$ $\displaystyle =$ $\displaystyle \alpha R$
The period of $\displaystyle \sin x$ and of $\displaystyle \cos x$ is $\displaystyle 2\pi$
$\displaystyle \sin(x+2\pi)$ $\displaystyle =$ $\displaystyle \sin x$
The period of $\displaystyle \tan x$ is $\displaystyle \pi$
$\displaystyle \tan(x+\pi)$ $\displaystyle =$ $\displaystyle \tan x$
The limit behind the derivative of $\displaystyle \sin x$ ( $\displaystyle x$ in radians)
$\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$

Trigonometric identities

The Pythagorean identity
$\displaystyle \sin^2\alpha+\cos^2\alpha$ $\displaystyle =$ $\displaystyle 1$
Tangent
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
Sine – an odd function
$\displaystyle \sin(-\alpha)$ $\displaystyle =$ $\displaystyle -\sin\alpha$
Cosine – an even function
$\displaystyle \cos(-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
The sine of the supplement to $\displaystyle \pi$
$\displaystyle \sin(\pi-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The cosine of the complement to $\displaystyle \frac{\pi}{2}$
$\displaystyle \cos\left(\frac{\pi}{2}-\alpha\right)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of a double angle
$\displaystyle \sin2\alpha$ $\displaystyle =$ $\displaystyle 2\sin\alpha\cdot\cos\alpha$
The cosine of a double angle
$\displaystyle \cos2\alpha=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$
The difference of two sines
$\displaystyle \sin\alpha-\sin\beta$ $\displaystyle =$ $\displaystyle 2\cos\frac{\alpha+\beta}{2}\cdot\sin\frac{\alpha-\beta}{2}$
The sine of a sum and of a difference
$\displaystyle \sin(\alpha\pm\beta)$ $\displaystyle =$ $\displaystyle \sin\alpha\cdot\cos\beta\pm\cos\alpha\cdot\sin\beta$

Derivatives of the trigonometric functions

The derivative of sine ( $\displaystyle x$ in radians)
$\displaystyle (\sin x)'$ $\displaystyle =$ $\displaystyle \cos x$
The derivative of cosine
$\displaystyle (\cos x)'$ $\displaystyle =$ $\displaystyle -\sin x$
The derivative of tangent
$\displaystyle (\tan x)'$ $\displaystyle =$ $\displaystyle \frac{1}{\cos^2x}$